• # question_answer 29) Balance the following equations in basic medium by ion-electron method and oxidation number method : (a) ${{P}_{4}}(s)+O{{H}^{-}}(aq)\to P{{H}_{3}}(g)HPO_{2}^{-}(aq)$ (b) ${{N}_{2}}{{H}_{4}}(l)+ClO_{3}^{-}(aq)\to P{{H}_{3}}(g)+HPO_{2}^{-}(aq)$ (c) ${{N}_{2}}{{H}_{4}}(l)+ClO_{3}^{-}(aq)\to P{{H}_{3}}(g)+HPO_{2}^{-}(aq)$ $+{{O}_{2}}(g)+{{H}^{+}}$

(a) Given equation is : ${{P}_{4}}+O{{H}^{-}}\to P{{H}_{3}}+HPO_{2}^{-}$ half equations are : ${{P}_{4}}\to P{{H}_{3}}$ ${{P}_{4}}\to HPO_{2}^{-}$ These half equations can be balanced as $[{{P}_{4}}+12{{H}^{+}}+12{{e}^{-}}\to 4P{{H}_{3}}]\times 2$ $On\,addition\frac{\frac{[{{P}_{4}}+8{{H}_{2}}O\to 4HPO_{2}^{-}+12{{H}^{+}}+8{{e}^{-}}]\times 3}{5{{P}_{4}}+24{{H}_{2}}O\to 8P{{H}_{3}}+12HPO_{2}^{-}+12{{H}^{+}}}}{{}}$converting above equation into basic medium we get $5{{P}_{4}}+12{{H}_{2}}O+12O{{H}^{-}}\to 8P{{H}_{3}}+12HPO_{2}^{-}$ (b) Given equation is : ${{N}_{2}}{{H}_{4}}(l)+ClO_{3}^{-}(aq)\to NO(q)+C{{l}^{-}}(aq)$ Half equations are: ${{N}_{2}}{{H}_{4}}(l)\to NO(q)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.......(i)$ ${{N}_{2}}{{H}_{4}}(l)\to NO(q)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.......(i)$ These half equations can be balanced as $[{{N}_{2}}{{H}_{4}}(l)]+2{{H}_{2}}O\to 2NO+8{{H}^{+}}+8{{e}^{-}}]\times 3$ On $ONaddition\frac{[ClO_{3}^{-}(aq)]+6{{H}^{+}}+6{{e}^{-}}\to C{{l}^{-}}(aq)+3{{H}_{2}}O]\times 4}{\frac{3{{N}_{2}}{{H}_{4}}(l)+4ClO_{3}^{-}(aq)\to 6NO+4C{{l}^{-}}(aq)+6{{H}_{2}}O}{{}}}$(c) $C{{l}_{2}}{{O}_{7}}(g)+{{H}_{2}}{{O}_{2}}(aq)\to ClO_{2}^{-}(aq)$ $+{{O}_{2}}(g)+{{H}^{+}}(aq)$ Step I: Splitting the equation into two half equations: $C{{l}_{2}}{{O}_{7}}(g)\to ClO_{2}^{-};$ ${{H}_{2}}{{O}_{2}}(aq)\to {{O}_{2}}(g)+{{H}^{+}}(aq)$ Step II: Balancing the element other than hydrogen and oxygen: $C{{l}_{2}}{{O}_{7}}(g)\to ClO_{2}^{-};$ ${{H}_{2}}{{O}_{2}}(aq)\to {{O}_{2}}(g)+{{H}^{+}}(aq)$ Step III: Balancing oxygen by adding water molecules: $C{{l}_{2}}{{O}_{7}}(g)\to ClO_{2}^{-}+3{{H}_{2}}O$ ${{H}_{2}}{{O}_{2}}(aq)\to {{O}_{2}}(g)+{{H}^{+}}(aq)$ Step IV: Balancing hydrogen by adding ${{H}^{+}}$ ions. $C{{l}_{2}}{{O}_{7}}+6{{H}^{+}}+8{{e}^{-}}\to 2ClO_{2}^{-}+3{{H}_{2}}O$ ${{H}_{2}}{{O}_{2}}(aq)\to {{O}_{2}}(g)+2{{H}^{+}}$ Step V: Balancing charge by adding electron, we get $C{{l}_{2}}{{O}_{7}}(g)+6{{H}^{+}}+8{{e}^{-}}\to 2ClO_{2}^{-}+3{{H}_{2}}O$ $[{{H}_{2}}O(aq)\to {{O}_{2}}(g)+2{{H}^{+}}+2{{e}^{-}}]\times 4$ Step VI: Adding above equations, we get $C{{l}_{2}}{{O}_{7}}(g)+4{{H}_{2}}{{O}_{2}}(aq)\to 2ClO_{2}^{-}+3{{H}_{2}}O+2{{H}^{+}}$ This equation can be balanced in basic medium by adding two $O{{H}^{-}}$ ions on both sides: $C{{l}_{2}}{{O}_{7}}(g)+4{{H}_{2}}{{O}_{2}}(aq)2O{{H}^{-}}\to 2ClO_{2}^{-}+$ $3{{H}_{2}}O+2{{H}^{+}}+2O{{H}^{-}}$ OR $C{{l}_{2}}{{O}_{7}}(g)+4{{H}_{2}}{{O}_{2}}(aq)2O{{H}^{-}}(aq)\to 2ClO_{2}^{-}+5{{H}_{2}}O$